Question

A positive charge q is placed in front of a conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube to the charges appearing on its surface.

Answer 1

Answer:

9/[4pivaresilon_od^2towards the charge q]

Explanation:

Answer 2

$E\ (Electric\ Field) = \frac {1}{4 \pi \varepsilon _0} \times {q}{d^2}$

Explanation:

We know that,

• Electric Field at a Point due to a given charge  

E = $\frac {Kq}{r^2}$

where,

• q = charge,

• r = distance between the point and the charge

So, E (Electric Field) = $\frac {1}{4 \pi \varepsilon _0} \times {q}{d^2}$

• As r = d,  

Hence, The electric field at the centre of the cube due to the charges appearing on its surface is  

$E\ (Electric\ Field) = \frac {1}{4 \pi \varepsilon _0} \times {q}{d^2}$